General Chemistry
General Chemistry
7th Edition
ISBN: 9780073402758
Author: Chang, Raymond/ Goldsby
Publisher: McGraw-Hill College
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Chapter 18, Problem 18.80QP
Interpretation Introduction

Interpretation:

The equilibrium constant Kp value should be derived given the oxygen removal reaction.

Concept Information:

Thermodynamics is the branch of science that relates heat and energy in a system.  The four laws of thermodynamics explain the fundamental quantities such as temperature, energy and randomness in a system.  Entropy is the measure of randomness in a system.  For a spontaneous process there is always a positive change in entropy.  Free energy (Gibbs free energy) is the term that is used to explain the total energy content in a thermodynamic system that can be converted into work.  The free energy is represented by the letter G.  All spontaneous process is associated with the decrease of free energy in the system. The equation given below will enable us find the spontaneity of a reaction.  The equation given below helps us to calculate the change in free energy in a system.

  ΔG = ΔΗ- TΔS

Where,

  ΔG  is the change in free energy of the system

  ΔΗ is the change in enthalpy of the system

  T is the absolute value of the temperature

  ΔS is the change in entropy in the system

Equilibrium constant: The state in which the reactants and products have no net change over time. This is when the forward and reverse reactions occur at equal rates, this state of equilibrium can be described by the equilibrium constant (K). 

Free energy (Gibbs free energy) is the term that is used to explain the total energy content in a thermodynamic system that can be converted into work.  The free energy is represented by the letter G.  All spontaneous process is associated with the decrease of free energy in the system.  The standard free energy change (ΔG°rxn) is the difference in free energy of the reactants and products in their standard state.

ΔG°=-RTlnΔG=Free energyΔG°=Standard state free energyR=GasConstant(0.0826l.atm/K.atm)T=Temprature273KK=EqulibriumConstant(KPandKC)ln=(ve(log)State Function)ΔG°rxn=nΔGf°(Products)-nΔGf°(Reactants)

Where, "n" is the number of moles

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Chapter 18 Solutions

General Chemistry

Ch. 18.6 - Practice Exercise Calculate the equilibrium...Ch. 18.6 - Prob. 2PECh. 18.6 - Prob. 3PECh. 18.6 - Prob. 1RCCh. 18 - Prob. 18.1QPCh. 18 - Prob. 18.2QPCh. 18 - Prob. 18.3QPCh. 18 - Prob. 18.4QPCh. 18 - Prob. 18.5QPCh. 18 - Prob. 18.7QPCh. 18 - Prob. 18.8QPCh. 18 - Prob. 18.9QPCh. 18 - 18.10 Arrange the following substances (1 mole...Ch. 18 - Prob. 18.11QPCh. 18 - Prob. 18.12QPCh. 18 - Prob. 18.13QPCh. 18 - 18.14 State whether the sign of the entropy...Ch. 18 - 18.15 Define free energy. What are its units? Ch. 18 - 18.16 Why is it more convenient to predict the...Ch. 18 - 18.17 Calculate ΔG° for the following reactions at...Ch. 18 - 18.18 Calculate ΔG° for the following reactions at...Ch. 18 - Prob. 18.19QPCh. 18 - Prob. 18.20QPCh. 18 - Prob. 18.21QPCh. 18 - Prob. 18.22QPCh. 18 - Prob. 18.23QPCh. 18 - 18.24 For the autoionization of water at...Ch. 18 - Prob. 18.25QPCh. 18 - Prob. 18.26QPCh. 18 - Prob. 18.27QPCh. 18 - Prob. 18.28QPCh. 18 - Prob. 18.29QPCh. 18 - Prob. 18.30QPCh. 18 - Prob. 18.31QPCh. 18 - Prob. 18.32QPCh. 18 - Prob. 18.33QPCh. 18 - Prob. 18.34QPCh. 18 - Prob. 18.35QPCh. 18 - Prob. 18.36QPCh. 18 - Prob. 18.37QPCh. 18 - Prob. 18.38QPCh. 18 - Prob. 18.39QPCh. 18 - Prob. 18.40QPCh. 18 - Prob. 18.41QPCh. 18 - Prob. 18.42QPCh. 18 - Prob. 18.43QPCh. 18 - Prob. 18.44QPCh. 18 - Prob. 18.45QPCh. 18 - Prob. 18.46QPCh. 18 - 18.47 Calculate the equilibrium pressure of CO2...Ch. 18 - Prob. 18.48QPCh. 18 - 18.49 Referring to Problem 18.48, explain why the...Ch. 18 - Prob. 18.50QPCh. 18 - Prob. 18.51QPCh. 18 - Prob. 18.52QPCh. 18 - Prob. 18.53QPCh. 18 - Prob. 18.54QPCh. 18 - Prob. 18.55QPCh. 18 - 18.56 Crystallization of sodium acetate from a...Ch. 18 - Prob. 18.57QPCh. 18 - Prob. 18.58QPCh. 18 - Prob. 18.59QPCh. 18 - Prob. 18.60QPCh. 18 - Prob. 18.61QPCh. 18 - Prob. 18.62QPCh. 18 - Prob. 18.63QPCh. 18 - Prob. 18.64QPCh. 18 - Prob. 18.65QPCh. 18 - Prob. 18.66QPCh. 18 - Prob. 18.67QPCh. 18 - Prob. 18.68QPCh. 18 - Prob. 18.69QPCh. 18 - Prob. 18.70QPCh. 18 - Prob. 18.71QPCh. 18 - Prob. 18.72QPCh. 18 - 18.73 (a) Over the years there have been numerous...Ch. 18 - Prob. 18.74QPCh. 18 - 18.75 Shown here are the thermodynamic data for...Ch. 18 - Prob. 18.76QPCh. 18 - Prob. 18.77QPCh. 18 - Prob. 18.78QPCh. 18 - Prob. 18.79QPCh. 18 - Prob. 18.80QPCh. 18 - Prob. 18.81QPCh. 18 - Prob. 18.82QPCh. 18 - Prob. 18.83QPCh. 18 - 18.84 Large quantities of hydrogen are needed for...Ch. 18 - Prob. 18.85QPCh. 18 - Prob. 18.86QPCh. 18 - Prob. 18.87QPCh. 18 - Prob. 18.88QPCh. 18 - Prob. 18.89QPCh. 18 - Prob. 18.90QPCh. 18 - Prob. 18.91QPCh. 18 - Prob. 18.92QPCh. 18 - Prob. 18.93QPCh. 18 - Prob. 18.94QPCh. 18 - Prob. 18.95QPCh. 18 - Prob. 18.96QPCh. 18 - Prob. 18.98QPCh. 18 - Prob. 18.100SPCh. 18 - Prob. 18.101SPCh. 18 - Prob. 18.102SPCh. 18 - Prob. 18.103SPCh. 18 - Prob. 18.104SPCh. 18 - Prob. 18.105SPCh. 18 - Prob. 18.106SPCh. 18 - Prob. 18.107SPCh. 18 - Prob. 18.108SPCh. 18 - 18.109 The boiling point of diethyl ether is...
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